Question: Simplify the following expression: $\dfrac{72y^4}{48y^3}$ You can assume $y \neq 0$.
Explanation: $ \dfrac{72y^4}{48y^3} = \dfrac{72}{48} \cdot \dfrac{y^4}{y^3} $ To simplify $\frac{72}{48}$ , find the greatest common factor (GCD) of $72$ and $48$ $72 = 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3$ $48 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3$ $ \mbox{GCD}(72, 48) = 2 \cdot 2 \cdot 2 \cdot 3 = 24 $ $ \dfrac{72}{48} \cdot \dfrac{y^4}{y^3} = \dfrac{24 \cdot 3}{24 \cdot 2} \cdot \dfrac{y^4}{y^3} $ $\phantom{ \dfrac{72}{48} \cdot \dfrac{4}{3}} = \dfrac{3}{2} \cdot \dfrac{y^4}{y^3} $ $ \dfrac{y^4}{y^3} = \dfrac{y \cdot y \cdot y \cdot y}{y \cdot y \cdot y} = y $ $ \dfrac{3}{2} \cdot y = \dfrac{3y}{2} $